Question 18 of 58intermediate🔧 ApplyLong Answer2 marks

Verify A(adj A) = (adj A)A = |A|I:

[23−4−6]\begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}
Correct Answer

Given matrix A = [23−4−6]\begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}

Step 1: Find |A| |A| = (2)(-6) - (3)(-4) = -12 + 12 = 0

Step 2: Find adj A For a 2×22 \times 2 matrix, adj A = [a22−a12−a21a11]\begin{bmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{bmatrix} So, adj A = [−6−342]\begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix}

Step 3: Find A(adj A) A(adj A) = [23−4−6][−6−342]=[−12+12−6+624−2412−12]=[0000]\begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} -12+12 & -6+6 \\ 24-24 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

Step 4: Find (adj A)A (adj A)A = [−6−342][23−4−6]=[−12+12−18+188−812−12]=[0000]\begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} = \begin{bmatrix} -12+12 & -18+18 \\ 8-8 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

Step 5: Find |A|I |A|I = 0 × [1001]=[0000]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

Hence, A(adj A) = (adj A)A = |A|I is verified.

Exercise: EXERCISE 4.4 | Q: 3 | (Page 17)
For More Understanding

Explanation

This question verifies Theorem 1 from the textbook which states that for any square matrix A, A(adj A) = (adj A)A = |A|I. The key steps involve calculating the determinant, finding the adjoint using the shortcut formula for 2×22 \times 2 matrices, and then performing matrix multiplication. Since |A| = 0, both products result in a zero matrix, which equals |A|I.

Solution Steps

  1. Step 1: Calculate determinant |A| = 0

  2. Step 2: Find adj A using the formula for 2×22 \times 2 matrices

  3. Step 3: Compute A(adj A) = zero matrix

  4. Step 4: Compute (adj A)A = zero matrix

  5. Step 5: Verify |A|I = zero matrix, confirming the theorem

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